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(D) For both cylinders $A$ and $B$,the gas is diatomic,so the adiabatic index $\gamma = 1.4$.For cylinder $A$,the piston is free,so the process is iso

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$42$

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(D) For both cylinders $A$ and $B$,the gas is diatomic,so the adiabatic index $\gamma = 1.4$.For cylinder $A$,the piston is free,so the process is isobaric (constant pressure). The heat supplied is $\Delta Q = n C_P \Delta T_A$.For cylinder $B$,the piston is fixed,so the process is isochoric (constant volume). The heat supplied is $\Delta Q = n C_V \Delta T_B$.Since the heat supplied is the same for both,we have:$n C_P \Delta T_A = n C_V \Delta T_B$Dividing both sides by $n C_V$,we get:$\Delta T_B = \frac{C_P}{C_V} \Delta T_A$Since $\frac{C_P}{C_V} = \gamma$,we have:$\Delta T_B = \gamma \Delta T_A$Given $\gamma = 1.4$ and $\Delta T_A = 30 \ K$:$\Delta T_B = 1.4 	imes 30 \ K = 42 \ K$.

List-$I$	List-$II$
$A$. Zeroth law of thermodynamics	$I$. Direction of flow of heat
$B$. First law of thermodynamics	$II$. Work done is zero
$C$. Free expansion of a gas	$III$. Thermal equilibrium
$D$. Second law of thermodynamics	$IV$. Law of conservation of energy

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